问题描述
我有一个Num_tuples元组的列表，它们都具有相同的长度Dim_tuple
I have a list of Num_tuples tuples that all have the same length Dim_tuple
xlist = [tuple_1, tuple_2, ..., tuple_Num_tuples]
为了明确起见，假设Num_tuples=3和Dim_tuple=2
For definiteness, let's say Num_tuples=3 and Dim_tuple=2
xlist = [(1, 1.1), (2, 1.2), (3, 1.3)]
我想使用用户提供的列名user_names列表和用户提供的变量类型user_types
I want to convert xlist into a structured numpy array xarr using a user-provided list of column names user_names and a user-provided list of variable types user_types
user_names = [name_1, name_2, ..., name_Dim_tuple]
user_types = [type_1, type_2, ..., type_Dim_tuple]
因此在创建numpy数组时，
So in the creation of the numpy array,
dtype = [(name_1,type_1), (name_2,type_2), ..., (name_Dim_tuple, type_Dim_tuple)]
在我的玩具示例中，所需的最终产品看起来像这样:
In the case of my toy example desired end product would look something like:
xarr['name1']=np.array([1,2,3])
xarr['name2']=np.array([1.1,1.2,1.3])
我如何切片xlist来创建xarr而没有任何循环?
How can I slice xlist to create xarr without any loops?
推荐答案
元组列表是向结构化数组提供数据的正确方法:
A list of tuples is the correct way of providing data to a structured array:
In [273]: xlist = [(1, 1.1), (2, 1.2), (3, 1.3)]
In [274]: dt=np.dtype('int,float')
In [275]: np.array(xlist,dtype=dt)
Out[275]:
array([(1, 1.1), (2, 1.2), (3, 1.3)],
dtype=[('f0', '<i4'), ('f1', '<f8')])
In [276]: xarr = np.array(xlist,dtype=dt)
In [277]: xarr['f0']
Out[277]: array([1, 2, 3])
In [278]: xarr['f1']
Out[278]: array([ 1.1, 1.2, 1.3])
或者名称是否重要:
In [280]: xarr.dtype.names=['name1','name2']
In [281]: xarr
Out[281]:
array([(1, 1.1), (2, 1.2), (3, 1.3)],
dtype=[('name1', '<i4'), ('name2', '<f8')])
http://docs.scipy. org/doc/numpy/user/basics.rec.html#filling-structured-arrays
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